/*
Considering 4-digit primes containing repeated digits it is clear that they cannot all be the same: 1111 is divisible by 11, 2222 is divisible by 22, and so on. But there are nine 4-digit primes containing three ones:
1117, 1151, 1171, 1181, 1511, 1811, 2111, 4111, 8111
We shall say that M(n, d) represents the maximum number of repeated digits for an n-digit prime where d is the repeated digit, N(n, d) represents the number of such primes, and S(n, d) represents the sum of these primes.
So M(4, 1) = 3 is the maximum number of repeated digits for a 4-digit prime where one is the repeated digit, there are N(4, 1) = 9 such primes, and the sum of these primes is S(4, 1) = 22275. It turns out that for d = 0, it is only possible to have M(4, 0) = 2 repeated digits, but there are N(4, 0) = 13 such cases.
In the same way we obtain the following results for 4-digit primes.

Digit, d
M(4, d)
N(4, d)
S(4, d)

0
2
13
67061

1
3
9
22275

2
3
1
2221

3
3
12
46214

4
3
2
8888

5
3
1
5557

6
3
1
6661

7
3
9
57863

8
3
1
8887

9
3
7
48073


For d = 0 to 9, the sum of all S(4, d) is 273700.
Find the sum of all S(10, d).

Anser:
Time:
*/
package main

import (
	"fmt"
	"time"
)

func main() {
	tstart := time.Now()



	tend := time.Now()
	fmt.Println(tend.Sub(tstart))
}